cooked proofs q (1 Viewer)

v.tex · Sunday at 8:49 PM

killer queen said:
is that as simple as I think it is

maybe the solution is simple

cheesynooby · Sunday at 8:54 PM

v.tex said:
View attachment 46912If you can't do this ur cooked

how

cheesynooby · Sunday at 8:57 PM

nvm but that took me way longer than it shouldve

Daq · Monday at 2:54 PM

By AM-GM inequality,

$x^5+y^5 \ge 2\sqrt{x^5y^5}$
$x^5+y^5 \ge 2x^2y^2\sqrt{xy}$
$x^5+y^5\ge x^2y^2(x+y) \text{ (By AM-GM inequality)}$
$x^5+y^5\ge xy(xy)(2) \text{ as }x,y \ge 1$
$x^5+y^5\ge xy\left(\frac{x^2+y^2}{2}\right)(2) \text{ (By AM-GM inequality)$
$x^5+y^5 \ge xy(x^2+y^2)$

Edit* This solution is wrong, because $\frac{a+b}{2} \ge \sqrt{ab}$ , so if you sub it in to the inequality, it is wrong

C2H6O · Monday at 3:45 PM

damn these solutions are so creamy wtf

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cooked proofs q (1 Viewer)

v.tex

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cheesynooby

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cheesynooby

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Daq

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